Atyendix VL
1097
perpendicular to the track. Find (i) the speed of the car, (2) the
coefficient of friction if the car is just on the point of side-slipping.
(Board of Education Exam. Stage 3, 1904.)
Referring to Fig. 968. There are two forces, the weight of the car
and the centrifugal force, and the resultant of these takes the same
direction as the plumb-bob string, for their ratio is independent of
weight, viz. :-
tan 22*
2/2 _ tan 22°
And V = ^.4.04 x 32 x 300 = 62*2 ft. per sec.
or 42*4 miles per hr.
Also the angle of resultant with the track = 12°
.". fji = tan 12° = '208.
Example 76. A Trarncar weighs 50 tons, and current is cut oft
when the speed is 16 miles per hr. Reckoning" time from that instant,
the following velocities and times were noted:
V (rniles per hr,) ... ......... 16 14 12 10
/ (time in sees.)............ o 9*3 21 35
Calculate the average retarding force, and find the average-velocity
from t = o to t = 35, Find distance travelled between these times.
"If the law of resistance be F Ibs, = a 4- b V -f <; V2, find the values
of a, t>, and<r, from the above observations. (Hons. Applied Mech.
Exam.)
Set out the velocities on a time base, Fig. 9680.
area under V curve.' _ 129-5-+152-1 + J54
Average velocity - -......~Stffltoc""......... ~ ^
total time
= iŁ44rniles Per
= 50 x 224.0
Ibs. - f / -
32
7 - 5ioo?
where V is the decrease of velocity ia miles per hr.
Hence average force,,* '
874 Ibs.